Answer: B
\[ \tan\theta = \sqrt{3} = \tan\left(\frac{\pi}{3}\right) \Rightarrow \theta = \frac{\pi}{3} \] \[ \text{Since } \theta + \phi = \frac{7\pi}{12} \Rightarrow \frac{\pi}{3} + \phi = \frac{7\pi}{12} \Rightarrow \phi = \frac{7\pi}{12} - \frac{\pi}{3} \Rightarrow \phi = \frac{7\pi - 4\pi}{12} = \frac{3\pi}{12} = \frac{3 \times 180^\circ}{12} = 45^\circ \] \[ \therefore \tan\phi = \tan 45^\circ = 1 \]
\[ \tan\theta = \sqrt{3} = \tan\left(\frac{\pi}{3}\right) \Rightarrow \theta = \frac{\pi}{3} \] \[ \text{Since } \theta + \phi = \frac{7\pi}{12} \Rightarrow \frac{\pi}{3} + \phi = \frac{7\pi}{12} \Rightarrow \phi = \frac{7\pi}{12} - \frac{\pi}{3} \Rightarrow \phi = \frac{7\pi - 4\pi}{12} = \frac{3\pi}{12} = \frac{3 \times 180^\circ}{12} = 45^\circ \] \[ \therefore \tan\phi = \tan 45^\circ = 1 \]